Worked out
ANSWERS
to the specific heat capacity questions Q1
A 0.5 kg block of copper absorbed 1520 J of energy and its temperature rose by
8.0^{o}C.
Calculate the specific heat capacity of
copper.
∆E = m x c x ∆T
SHC_{copper} = ∆E / (m x ∆T)
SHC_{copper} = ∆E / (m x ∆T)
= 1520 / (0.5 x 8.0) =
380 J/kg^{o}C
Q2
A block bronze has a specific heat capacity of 400 J/kg^{o}C.
If a 1500 g block of bronze absorbs 3000
J of energy, what would be the rise in its temperature?
∆E = m x c x ∆T, 1500 g
= 1.500 kg
so rearranging: ∆T = ∆E /
(m x SHC_{bronze}) = 3000 / (1.5 x 400) =
5.0^{o}C
Q3
Calculate the thermal energy in kJ required to heat 2 kg of water (SHC = 4180
J/kg^{o}C) from 20^{o}C to 100^{o}C.
∆E = m x c x ∆T = 2 x 4180 x
(10020) = 6.68 x 10^{5} J
Thermal energy required =
668 kJ
Q4 A storage tank of water contains 500 kg of
hot water at 80^{o}C. (SHC water = 4180 J/kg^{o}C)
How many MJ of thermal energy can be
transferred to a heating system if the temperature of the water falls to 25^{o}C?
∆E = m x c x ∆T = 500 x 4180
x (8025) = 1.15 x 10^{8} J
Thermal energy transferred = 1.15 x 10^{8}
/ 10^{6} = 115 MJ
(1.15 x 10^{2} MJ, 3 sf)
Q5 A more complicated problem to solve  so
'double' think carefully!
Suppose 2 kg of steel at a temperature of
80^{o}C is placed in 10 kg (~10 litres) of water at 15^{o}C.
Calculate the final temperature of the
materials.
SHC steel = 450 J/kg^{o}C,
SHC water = 4180 J/kg^{o}C
The loss in J from the steel energy
store = gain in J of the thermal energy store of the water.
At 'thermal equilibrium' they will be
both at the same temperature, lets call this T.
The temperature fall of the steel is (80
 T), the temperature rise of the water is (T  15)
The loss from the steel thermal energy
store = ∆E = m x c x ∆T = 2 x 450 x (80  T) = 900 x (80  T) J
The gain of the water thermal energy
store = ∆E = m x c x ∆T = 10 x 4180 x (T  15) = 41800 x
(T  15) J
So, change in steel thermal energy store
= change in water thermal energy store
900 x (80  T) = 41800 x (T 
15)
72000  900T = 41800T  627000
(changing sides, changing signs, we
get)
72000 + 627000 = 900T + 41800T
699000 = 42700T
T = 699000 / 42700 =
16.4^{o}C
(3 sf)
Q6 A night storage heater contains 75 kg of
concrete (SHC = 800 J/kg^{o}C)
At night when electricity is cheaper and
the house cooler, how many kJ of thermal energy is needed to raise the
temperature of the concrete from 8^{o}C to 35^{o}C.
∆E = m x c x ∆T = 75 x
800 x (35  8) = 1.62 x 10^{6} J
1.62 x 10^{6} / 1000 =
1620 J
Q7
This question is based on a physics experiment I did in school back in circa
1961.
Its a good little experiment to do.
Accurately weigh 100.0 g (~100 ml) of
water into a beaker at room temperature.
A 50.0 g brass weight was held by a thin
wire string or with a pair tongs in the centre of a roaring Bunsen flame.
After a few minutes the brass weight was
plunged into a beaker of 100 g cold water whose initial temperature was 19.5^{o}C.
After gently stirring carefully with a
thermometer the temperature of the water rose to 55.5^{o}C.
SHC of brass = 380 J/kg^{o}C, SHC
water = 4180 J/kg^{o}C
(a) If the flame temperature is T,
calculate the two thermal energy store transfers.
Thermal energy transfer from brass
weight = ∆E = m x c x ∆T = (50/1000) x 380 x (T  19.5) = 19.0 (T 
19.5) J
Thermal energy transfer from brass
weight = ∆E = m x c x ∆T = (100/1000) x 4180 x (55.5  19.5) = 15048 J
(b) Calculate the flame temperature T
Thermal energy transfer from brass
weight (J) = thermal energy transfer to water (J)
19.0 (T  19.5) = 15048
19.0T  370.5 = 15048
19.0T = 15418.5
T = 15418.5 / 19.0 =
812^{o}C
(3 sf)
(b) Suggest some sources of error in the
experiment
(i) the weight loses some thermal
energy to air on transfer  possibly quite a bit, since the experimental
answer you get is often well below the real answer for the flame
temperature of a bunsen flame ~10001200^{o}C.
(ii) the glass beaker absorbs some
heat, therefore its heat capacity has been ignored
(iii) the lower part of the wire will
transfer a little heat  but if you use brass wire, you could weigh that
too and make sure everything is immersed in the water!
Q8
A car of mass 1000 kg is moving at 20.0 m/s is brought to a halt by sharp
braking.
Assume all the kinetic energy is
converted to thermal energy by four steel brake drums.
If the total mass of the steel drums of
the brakes is 25.0 kg, assuming no heat losses, calculate the maximum rise
in temperature the brake drums will attain.
(a) First calculate the kinetic energy
(KE) of the car.
KE = ½ mv^{2}
KE = ½ x 1000 x 20^{2} = 2.0
x 10^{5} J
(b) Calculate the temperature rise of the
braking system (SHC of steel = 450 J/kg^{o}C)
∆E = m x c x ∆T
∆T = ∆E / (m x c) = 2.0 x 10^{5}
/ (25 x 450) =
17.8^{o}C
(3sf)
The next two questions are more tricky
and involve using electricity formulae:
Electricity power calculations, P = IV
The specific heat capacity of water is
4180 J/kg^{o}C. Be careful with all the units.
Q9
A small electric kettle is connected to the 240 V AC mains supply and uses a
current of 8.0 A.
It contains 1.5 kg of water (~1.5 litres,
1500 ml, 1500 cm^{3}) at 20^{o}C.
(a) Calculate the power of the kettle and
the rate of energy transfer.
P (W) = I (a) x V (p.d. in volts) = 8
x 240 =
1920 W =
1920
J/s
(b) If the kettle is switched on for 2.0
minutes, how much energy is transferred to the thermal energy store of the
water?
P = 1920 J/s, energy transferred =
power x time = 1920 x 2 x 60 = 230400 J =
2.3
x 10^{5}
J
(c) What temperature will the hot water
rise too?
∆E = mass of water x SHC_{H2O} x temperature rise
∆E (J) = m (kg) x c (J/kgoC) x ∆T (^{o}C)
Rearranging: ∆T = ∆E / (m x SHC_{H2O})
∆T = 230400 / (1.5 x 4180) = 36.7 ^{o}C
Therefore temperature after 2 minutes
of heating is 20 + 36.7 =
56.7^{o}C
(d) How much energy is needed to raise
the water temperature from 20^{o}C to 100^{o}C?
∆E = mass of water x SHC_{H2O} x temperature rise
∆E = 1.5 x 4180 x (100  20) = 5.016 x 10^{5} J =
5.01 x 10^{5}
J (3 sf)
(e) Starting with the cold water at 20^{o}C,
how long would it take to boil the kettle' in minutes and seconds?
What assumptions have you made for
this calculation?
The kettle's power s 1920 W, that is
1920 J/s, and energy needed = 5.016 x 10^{5} J
P = E / t, so t = E / P = 5.016 x 10^{5}
/ 1920 = 261.25 s.
Time to boil =
4 mins and 21 seconds
(to the nearest second).
This calculation assumes all the
electrical energy converted to thermal energy, actually increased the
thermal energy store of the water.
However, you would always get a small
amount of wasted thermal energy transferring to the thermal energy store
of the kettle casing by conduction, and convection and radiation
transfer to the thermal energy store of he surrounding air.
Q10
A hot water cylinder (immersion heater) is quoted as having a
capacity of '120 litres of water' (120 kg of water).
The tank is fitted with a 3 kW heating
element working off 240 V mains electricity.
(a) How much energy is needed to heat
fresh cold water at 12^{o}C to a temperature of 70^{o}C?
∆E = mass of water x SHC_{H2O} x temperature rise
Energy needed = 120 x 4180 x (70 
12) = 29 092 800 J =
2.91 x 10^{7}
J (3 sf)
(b) How long will it take to heat the
water up? (to the nearest minute)
P (J/s) = E (J) / t (s), 3000 = 29
092 800 / t
time = E / P = 29 092 800 / 3000 =
9697.6 seconds
9697.6 / 60 = 161.6. time
required =
162 minutes (sf)
(c) What current is flowing through the
heating element?
P = I x V, I = P / V = 3000 /
240 = 12.5 A
(d) What is the resistance of the heating
element?
V = IR, R = V / I = 240 / 12.5
= 19.2 Ω
(e) If the immersion was fitted with a 6
kW heating element, how would this affect the time to heat water up to a
certain temperature?
This is twice the power of the 3 kW
heating element.
P = E / t, t = E / P, for the
same quantity of water and the same temperature rise, it would
take half the time
compared to the 3 kW heater.
Using SEARCH some initial results may be ad links you
can ignore  look for docbrown
INDEX for my physics notes on specific
heat capacity
